Source: Introduction to Mechanics Kleppner and Kolenkow:

The “mass” in the law of gravitation (gravitational mass) measures the strength of gravitational interaction and is operationally distinct from the “mass” (inertial mass) that characterizes inertia in Newton’s second law. Why gravitational mass is proportional to inertial mass is a deep mystery. Newton recognized the mystery and confirmed the fact experimentally to an accuracy of about 1% by observing that the period of a pendulum does not depend on the material of the pendulum.

What experiment is the author talking about? How did Newton find the relation between inertial mass and gravitational mass by knowing that period of pendulum doesn't depends upon material of pendulum? Couldn't just catch it.


For small oscillations, you find that $$T=2\pi\sqrt{\frac{L}{g}}$$, where $L$ is the string lenght, $g$ the gravity, and $T$ the period. You see that this doesn't depend on the mass.

But the key is the process of obtaining this. There are two forces: tension and weight:

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The tension must cancel out with the weight's component in that direction. The remaaining "horizontal" component of weight is the responsible of the oscillation.

This component of the weight: $m g \sin(\theta)$ is the only force (the other ones cancel each other).

So, using Newton's law, you'd get

$$m_I\cdot a = m_g\cdot g\cdot \sin(\theta)$$

Notice that "sum of forces = inertial mass times acceleration", so the left hand side has inertial mass $m_I$. However, the weight is given by "gravitational mass times gravity" $m_g\cdot g$.Then it's multipleid by the sine because we're taking its horizontal component.

If oscillations are small enough, we can make an approximation: $sin(\theta)\approx \theta$. This approximaton makes the problem much easier to solve, and it is very accurate.

So we have $$m_I\cdot a \approx m_g\cdot g \cdot \theta$$

If you replace $a=l \alpha$, where $\alpha$ is the angular acceleration, then we have

$$m_I\cdot l \cdot \alpha \approx m_g\cdot g \cdot \theta$$

which you can rearrange as

$$ \alpha\approx \frac{m_g}{m_I} \frac{g}{l} \theta$$

or equivalently as

$$ \ddot{\theta}- \frac{m_g}{m_I} \frac{g}{l} \theta \approx 0$$

since $\alpha=\ddot{\theta}$. This is the famous equation of the harmonic oscilaltor, with a frequency given by what is next to the angle:

$$\omega^2 = \frac{m_g}{m_I} \frac{g}{l}$$

Which implies that the period is

$$T=\frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m_I}{m_g} \frac{L}{g}} $$

The thing is that, experimentally, $m_I=m_g$ is found, and hence the period depends only on the lenght, and not on the gravitational mass we add on it.

  • $\begingroup$ This shows inertial mass equals passive gravitational mass, which is not so mysterious now after we have GR. But why active gravitational mass equals passive gravitational mass equals inertial mass? $\endgroup$ – velut luna Mar 25 '18 at 10:09
  • $\begingroup$ How was it experimentally proven that, mI=mg? Eötvös experiment? Was it done at the time of Newton? $\endgroup$ – suiz Mar 25 '18 at 10:30
  • $\begingroup$ I've got the answer from the link mathpages.com/home/kmath582/kmath582.htm . $\endgroup$ – suiz Mar 28 '18 at 6:28

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